Simplify the following expression and state the condition under which the simplification is valid. You can assume that $n \neq 0$. $x = \dfrac{8n}{8n + 8} \times \dfrac{7(n + 1)}{2} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $x = \dfrac{ 8n \times 7(n + 1) } { (8n + 8) \times 2 } $ $ x = \dfrac {8n \times 7(n + 1)} {2 \times 8(n + 1)} $ $ x = \dfrac{56n(n + 1)}{16(n + 1)} $ We can cancel the $n + 1$ so long as $n + 1 \neq 0$ Therefore $n \neq -1$ $x = \dfrac{56n \cancel{(n + 1})}{16 \cancel{(n + 1)}} = \dfrac{56n}{16} = \dfrac{7n}{2} $